Question

Three charges q, 2q and q are to be placed on a straight wire of length 10 m.
Determine the positions where the charges should to be placed so that the
electrostatic potential energy of the system is a minimum.

Answer 1

Answer on Question #86422, Physics / Electromagnetism

Let the potential be 0 at infinity. Let's make an assumption that the 2q and q charges are at the opposite end of the line, this would be the minimum potential energy positions if the q charge was not included, and switching the q charge with either of the larger charges would only increase the potential energy.

Let $r$ be the distance of the $q$ charge from the 2q charge. Then the potential energy of the system is:

$$U = K \left( \frac{2q^2}{r} + \frac{q^2}{10-r} + \frac{2q^2}{10} \right)$$

We can minimize this by taking the derivative with respect to $r$ and setting it equal to 0:

$$0 = k q^2 \left( \frac{-2}{r^2} + \frac{1}{(10-r)^2} \right)$$ $$\frac{r^2}{2} = (10 - r)^2$$ $$r = 20 \pm 10\sqrt{2}$$

This has 2 solutions. However, we are only interested in solutions that lie along the $10\mathrm{cm}$ line (solutions where $r$ is positive), so

$$r = 20 - 10\sqrt{2}$$

Which is indeed on the line and fulfils our intuition that the q charge should be closer to the 2q charge than the q charge.

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